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When we collect data from an experiment, it can be “distributed”
(spread out) in different ways.
There are many cases where the data tends to be around a central
value, equally spread left and right, getting close to a
“Normal Distribution.”
The yellow histogram shows data that follows
closely, not necessarily perfectly, a bell curve.
The Normal Distribution has:
• Mean = Median = Mode
• Symmetry about the center
• 50% of values smaller than the mean and 50% of values greater
than the mean
Remember that the Standard Deviation is a measure of how spread
out numbers are for a set of data. After you compute the standard
deviation for a Normal Distribution, you will find out:
In summary (Empirical Rule):
99.7%
We can say that in a Normal Distribution any value is:
• Likely to be within 1 standard deviation of the mean (68%)
• Very likely to be within 2 standard deviations of the mean (95%)
• Almost certain to be within 3 standard deviations of the mean (99.7%)
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
1.1
1.7
The mean is exactly half way between 1.1 m and 1.7 m:
1.1 + 1.7
𝑥=
= 1.4 𝑚
2
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
1.1
1.40
1.7
The mean is exactly half way between 1.1 m and 1.7 m:
1.1 + 1.7
𝑥=
= 1.4 𝑚
2
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
2) Find the standard deviation for the students’ heights
1.1
1.40
1.7
A total of 4 standard deviations are between 1.1 m and 1.7 m:
1.7 − 1.1
𝜎=
= 0.15 𝑚
4
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
2) Find the standard deviation for the students’ heights
1.1 1.25 1.40 1.55 1.7
A total of 4 standard deviations are between 1.1 m and 1.7 m:
1.7 − 1.1
𝜎=
= 0.15 𝑚
4
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
2) Find the standard deviation for the students’ heights
3) What percent of students can we expect to have heights less
than 1.25 m? How many LHS students are shorter than 1.25 m?
3) 𝑃 ℎ < 1.25𝑚 = 16%
1800 × 0.16 = 288 students
0.95 1.1
1.25 1.4
1.55 1.7 1.85
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
2) Find the standard deviation for the students’ heights
3) What percent of students can we expect to have heights less
than 1.25 m? How many LHS students are shorter than 1.25 m?
4) What percent of students can we expect to be taller than 1.7 m?
How many LHS students are taller than 1.7 m?
4) 𝑃 ℎ > 1.7𝑚 = 2.5%
1800 × 0.025 = 45 students
0.95 1.1
1.25 1.4
1.55 1.7 1.85
Example: At Liberty, 95% of students are between 1.1 m and 1.7 m
tall. Assuming this data is normally distributed and there are 1,800
students at LHS,
1) Find the mean for the students’ heights
2) Find the standard deviation for the students’ heights
3) What percent of students can we expect to have heights less
than 1.25 m? How many students are shorter than 1.25 m?
4) What percent of students can we expect to be taller than 1.7 m?
How many LHS students are taller than 1.7 m?
5) What percent of students can we expect to have heights
between 0.95 m and 1.85 m? How many LHS students have
heights between 0.95 m and 1.85 m?
5) 𝑃 0.95 < ℎ < 1.85𝑚 ≈ 99.7%
1800 × 0.997 = 1794 students
0.95 1.1 1.25 1.4 1.55 1.7 1.85